c - 在没有显式类型的情况下评估 C 中的表达式

Question

考虑以前一篇关于 C 的论文中的以下问题:

Show step-by-step how the following C expression is evaluated, giving the value of the expression and the type of that value.

9 / 3 / 2 * 6 + 2 * 1.5

类型是 float 还是 double？有人可以澄清一下吗？ (请注意问题中缺少明确的类型。)

Answer 1

根据普通算术转换(6.3.1.8普通算术转换)

Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double.

答案:

1. 9/3 = 3(整数)
2. 3/2 = 1(整数)
3. 1 * 6 = 6(整数)
4. 2 * 1.5 = 3.0(双)
5. 6 + 3.0 = 9.0(双)